\[ \frac{16 \sqrt{3}}{\sqrt{8}} \] After you pick your answe

To simplify the expression \( \frac{16 \sqrt{3}}{\sqrt{8}} \), we start by simplifying \( \sqrt{8} \). We know \( \sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt{2} \). Now replace \( \sqrt{8} \) in the original expression: \[ \frac{16 \sqrt{3}}{\sqrt{8}} = \frac{16 \sqrt{3}}{2\sqrt{2}}. \] Next, divide the numerator by the denominator: \[ \frac{16}{2} = 8, \] so we rewrite the expression as: \[ 8 \cdot \frac{\sqrt{3}}{\sqrt{2}}. \] Now, we can rationalize it. Multiply the numerator and denominator by \( \sqrt{2} \): \[ 8 \cdot \frac{\sqrt{3} \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = 8 \cdot \frac{\sqrt{6}}{2} = 4\sqrt{6}. \] So the answer is \( 4\sqrt{6} \). To further enhance your understanding, did you know that simplifying radicals has been a part of mathematical education for centuries? Ancient Greeks were some of the first to explore the properties of square roots, laying the groundwork for today's algebraic concepts. It’s fascinating how these ideas have evolved! Moreover, in the real world, simplifying radicals is handy when solving geometry problems, especially when calculating the area or side lengths of triangles. For instance, if you know the lengths of sides in a right triangle, you may end up needing to simplify square roots to find the hypotenuse or perpendicular heights. It’s all about making those numbers easier to work with!